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Five fair coins are tossed one after the other. What is the probabilit
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06 Oct 2021, 09:15
06 Oct 2021, 09:15
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Five fair coins are tossed one after the other. What is the probability that only the first three are heads?
A. 1/32
B. 1/8
C. 5/16
D. 1/2
E. 11/16
A. 1/32
B. 1/8
C. 5/16
D. 1/2
E. 11/16
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Re: Five fair coins are tossed one after the other. What is the probabilit
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06 Oct 2021, 09:27
06 Oct 2021, 09:27
1
Carcass wrote:
Five fair coins are tossed one after the other. What is the probability that only the first three are heads?
A. 1/32
B. 1/8
C. 5/16
D. 1/2
E. 11/16
A. 1/32
B. 1/8
C. 5/16
D. 1/2
E. 11/16
P(HHHTT) = P(heads on toss 1 AND heads on toss 2 AND heads on toss 3 AND tails on toss 4 AND tails on toss 5)
= P(heads on toss 1) x P(heads on toss 2) x P(heads on toss 3) x P(tails on toss 4) x P(tails on toss 5)
= 1/2 x 1/2 x 1/2 x 1/2 x 1/2
= 1/32
Answer: A
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Re: Five fair coins are tossed one after the other. What is the probabilit
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18 Oct 2022, 09:49
18 Oct 2022, 09:49
1
Given that Five fair coins are tossed one after the other and we need to find What is the probability that only the first three are heads?
Let's solve the problem using two Methods:
Method 1:
We know that probability of getting a head, P(H), = Probability of getting a Tail, P(T) = \(\frac{1}{2}\)
=> P(HHHTT) = P(H) * P(H) * P(H) * P(T) * P(T) = \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) = \(\frac{1}{32}\)
So, Answer will be A
Method 2:
5 coins are tossed => Total number of cases = \(2^5\) = 32
Getting HHHTT is just one case out of these 32 cases
=> P(HHHTT) = \(\frac{1}{32}\)
So, Answer will be A
Hope it helps!
Watch the following video to learn How to Solve Probability with Coin Toss Problems
Let's solve the problem using two Methods:
Method 1:
We know that probability of getting a head, P(H), = Probability of getting a Tail, P(T) = \(\frac{1}{2}\)
=> P(HHHTT) = P(H) * P(H) * P(H) * P(T) * P(T) = \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) = \(\frac{1}{32}\)
So, Answer will be A
Method 2:
5 coins are tossed => Total number of cases = \(2^5\) = 32
Getting HHHTT is just one case out of these 32 cases
=> P(HHHTT) = \(\frac{1}{32}\)
So, Answer will be A
Hope it helps!
Watch the following video to learn How to Solve Probability with Coin Toss Problems
